Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

Body A of mass 4 m moving with speed u collides with another body B of mass 2 m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya

answer is 3.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

According to conservation of momentum,4mu1=4mv1+2mv2⇒ 2u1-v1=v2                          .........(i) From conversation of energy, 12(4m)u12=12(4m)v12+12(2m)v22 ⇒ 2u12-v12=v22                          ........(ii) From (i) and (ii) 2u12-v12=4u1-v12                    ..........(iii)  Now, fraction of loss in kinetic energy for mass 4m ,  ΔKKi=Ki-KfKi=1/2(4m)u12-12(4m)v1212(4m)u12             .........(iv)  Substituting (iii) in (iv), we get ΔKKi=89
Watch 3-min video & get full concept clarity

courses

No courses found

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon
Body A of mass 4 m moving with speed u collides with another body B of mass 2 m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is